Question

How do you find the vertical, horizontal or slant asymptotes for `y=(2x^2-3x+4)/(x+2)`?

Answer 1

Vertical asymptotes at `x_v = -2`
Slant asymptote `y = 2x-7`

Explanation:

In a polynomial fraction `f(x) = (p_n(x))/(p_m(x))` we have:

`1)` vertical asymptotes for `x_v` such that `p_m(x_v)=0`
`2)` horizontal asymptotes when `n le m`
`3)` slant asymptotes when `n = m + 1`
In the present case we have `x_v = -2` and `n = m+1` with `n = 2` and `m = 1`

Slant asymptotes are obtained considering #(p_n(x))/(p_{n-1}(x)) approx y = a+b x# for large values of `abs(x)`

In the present case we have

`(p_n(x))/(p_{n-1}(x)) = (2x^2-3x+4)/(x + 2)`
`p_n(x) = p_{n-1}(x)(a x + b) + r_{n-2}(x)`
`(p_n(x))/(p_{n-1}(x)) = (2x^2-3x+4) =(x+2) (a x +b)+c`

equating we have

#{ (4 - 2 b - c=0), (-3 - 2 a - b=0), (2 - a=0) :}#

Solving for `a,b,c` we have `a = 2,b=-7,c=18`
so the slant asymptote reads

`y = 2x-7`